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BITSAT 2009
Exam
BITSAT
Year
2009
Subject
Mathematics
Chapter
Application of Derivatives

BITSAT 2009 · Mathematics · Application of Derivatives

If (M(α)=[ array ccc α & - α & 0 α & α & 0 0 & 0 & 1 array ]);...

If (M(\alpha)=\left[cos⁡α−sin⁡α0sin⁡αcos⁡α0001\begin{array}{ccc}\cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1\end{array}\right]); (M(\beta)=\left[cos⁡β0sin⁡β010−sin⁡β0cos⁡β\begin{array}{ccc}\cos \beta & 0 & \sin \beta \\ 0 & 1 & 0 \\ -\sin \beta & 0 & \cos \beta\end{array}\right]) then ([M(\alpha) M(\beta)]^{-1}) is equal to -

Options

  1. A(\mathrm{M}(\beta) \mathrm{M}(\alpha))
  2. B(\mathrm{M}(-\alpha) \mathrm{M}(-\beta))
  3. C(\mathrm{M}(-\beta) \mathrm{M}(-\alpha))
  4. D(-\mathrm{M}(\beta) \mathrm{M}(\alpha))

Correct answer

C. (\mathrm{M}(-\beta) \mathrm{M}(-\alpha))

Step-by-step solution

([M(α)M(β)]−1=M(β)−1M(α)−1]Now M(α)−1⁡=[cos⁡αsin⁡α0−sin⁡αcos⁡α0001]=[cos⁡(−α)−sin⁡(−α)0sin⁡(−α)cos⁡(−α)0001]=M(−α)M(β)−1=[cos⁡β0−sin⁡β010sin⁡β0cos⁡β]=[cos⁡(−β)0sin⁡(−β)010−sin⁡(−β)0cos⁡(−β)]=M(−β)[M(α)M(β)]−1=M(−β)M(−α)\begin{aligned} & \left.[\mathrm{M}(\alpha) \mathrm{M}(\beta)]^{-1}=\mathrm{M}(\beta)^{-1} \mathrm{M}(\alpha)^{-1}\right] \\ & \operatorname{Now~M(\alpha )^{-1}}=\left[\begin{array}{ccc} \cos \alpha & \sin \alpha & 0 \\ -\sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{array}\right] \\ & =\left[\begin{array}{ccc} \cos (-\alpha) & -\sin (-\alpha) & 0 \\ \sin (-\alpha) & \cos (-\alpha) & 0 \\ 0 & 0 & 1 \end{array}\right]=\mathrm{M}(-\alpha) \\ & M(\beta)^{-1}=\left[\begin{array}{ccc} \cos \beta & 0 & -\sin \beta \\ 0 & 1 & 0 \\ \sin \beta & 0 & \cos \beta \end{array}\right] \\ & =\left[\begin{array}{ccc} \cos (-\beta) & 0 & \sin (-\beta) \\ 0 & 1 & 0 \\ -\sin (-\beta) & 0 & \cos (-\beta) \end{array}\right]=\mathrm{M}(-\beta) \\ & {[\mathrm{M}(\alpha) \mathrm{M}(\beta)]^{-1}=\mathrm{M}(-\beta) \mathrm{M}(-\alpha)} \\ & \end{aligned})
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